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diff --git a/944/CH3/EX3.21/example3_21_TACC.sce b/944/CH3/EX3.21/example3_21_TACC.sce
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+//example 3.21
+
+clear;
+clc;
+
+//Given:
+Cp1=poly([26.0, 43.5*10^-3, -148.3*10^-7],'T','c');//heat capacity for CO2(g)[J/K/mol]
+Cp2=poly([30.36, 9.61*10^-3, 11.8*10^-7],'T','c');//heat capacity for H2O(g)[J/K/mol]
+Cp3=poly([27.30, -5.23*10^-3, -0.04*10^-7],'T','c');//heat capacity for N2(g)[J/K/mol]
+H1=-881.25;//heat of combustion of methane at 298K[KJ/mol]:                CH4 + 2O2(g) --> CO2(g) + 2H2O(l)
+H2=43.6;//heat of vaporization of water at 298K[KJ/mol]:        H2O(l)-->H2O(g)  
+T1=298;//initial temperature [K]
+
+
+//to find the maximum flame temperature when one mole of methane is burnt completely  in calculated amount of air(N2 to O2 ratio 4),at constant pressure
+H=(H1+2*H2);//enthalpy of reaction: CH4 + 2O2(g) --> CO2(g) + 2H2O(g) [KJ]
+printf("H(enthalpy of reaction)=%f KJ/mol.\n",H);
+printf("H(enthalpy of heating)=%f KJ/mol. \n",-H);
+Cp=Cp1+2*Cp2+8*Cp3;//[J/K/mol]
+P=poly([0 , 305.12 , 52.28*10^-3 , -41.66*10^-7] , 'T' , 'c');
+i=horner(P,298);
+c1=-H*1000+i;
+P1=poly([-c1 , 305.12 , 52.28*10^-3 , -41.66*10^-7] , 'T' , 'c');
+T2=roots(P1)
+printf("T2 (maximum flame temperature)= %f K",T2(1));
+\ No newline at end of file